# Spacetime translation in QFT

I have a question about the field under the spacetime translation. For example, in page 26 of Peskin's textbook, they give the translation properties of the field. So consider the space translation, then $$e^{-i\vec{p} \cdot \vec{a}}\phi(\vec x)e^{i\vec{p} \cdot \vec a}=\phi({\vec x+\vec a})$$ It means that $$[\vec p,\phi(\vec x)]=i\frac{\partial}{\partial x}\phi(\vec x).$$ However in quantum mechanics, we know for any function $$A(x)$$, we have $$[\vec p,A(\vec x)]=-i\frac{\partial}{\partial x}A(\vec x).$$ So i am wondering why there is a minus sign missed in QFT.

## 1 Answer

Apart from the fact that your first equation is missing some $$i$$s in the exponential, the relative minus difference is just due to convention and it's not really a problem. I could have defined the transformation of the field operator (as some authors do) not as $$U^\dagger(a) \phi(x)U(a)$$ but as $$U(a)\phi(x)U^\dagger(a)$$ and this would have solved the minus sign problem. But as you can see, we have freedom in defining how the group action is on the field operators.

Moreover, in QFT, canonical commutators are not really a well defined things since field operators are distribution valued on all four variables, and this creates problems in interacting theories (which are the most relevant), but this is beyond the point.

• Thanks for your answer. So do you mean that the momentum operator in QM is different to the momentum operator in QFT, if i accept the convention in peskin's book Apr 7, 2021 at 2:50
• @chukk Yes, that's it. Apr 7, 2021 at 6:53